Physics Problems & Solutions: Magnetic Field

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Electromagnetism - Faraday’s law

The circuit shown is in a uniform magnetic field that is into the page and is decreasing in magnitude at rate of 150 tesla/second. The ammeter reads

A. 0.15 A
B. 0.35 A
C. 0.50 A
D. 0.65 A
E. 0.80 A

(GR9677 #02)

Solution:

V − IR − ɛ = 0
I = (V − ɛ)/R

ɛ = − dΦ/dt = −AdB/dt

Given:
dB/dt = −150 t/s (minus sign because it’s decreasing)
A = (0.1 m)²= 0.01 m²
R = 10 Ω
V = 5 V

ɛ = − (0.01)(−150) = 1.5 V
I = (5 − 1.5)/10 = 3.5/10 = 0.35 A

Answer: B

Nuclear & Particle Physics - Beta Decay

When the beta decay of ⁶⁰Co nuclei is observed at low temperature in a magnetic field that aligns spins of the nuclei, it is found that the electrons are emitted preferentially in a direction opposite to the ⁶⁰Co spin direction. Which of the following invariances is violated by this decay

A. Gauge invariance
B. Time invariance
C. Translation invariance
D. Reflection invariance
E. Rotation invariance

(GR9677 #34)

Solution:

Wu experiment, 1957: beta decay of Cobalt-60 shows that electrons are emitted preferentially in a direction opposite to the ⁶⁰Co spin direction → Parity violation by the weak interaction.

A. Gauge invariance → related to conservation of charge
B. Time invariance → energy
C. Translation invariance → momentum
D. Reflection invariance → parity
E. Rotation invariance → angular momentum

Answer: D

Electromagnetism - Faraday’s law

A circular wire loop of radius R rotates with an angular speed ω in a uniform magnetic field B, as shown in the figure. If the emf ɛ induced in the loop is ɛ₀ sin ωt, then the angular speed of the loop is

A. ɛ₀ R/B
B. 2πɛ₀/R
C. ɛ₀/BπR²
D. ɛ₀²/BR²
E. tan⁻¹(ɛ₀/Bc)

(GR9677 #46)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with Φ = NBA

Given:
N = 1
B uniform (constant)
ɛ = ɛ₀ sin ωt

→ ɛ = − B dA/dt = ɛ₀ sin ωt

− B dA = ɛ₀ sin ωt dt

B ∫dA = ɛ₀ ∫ sin ωt dt

− BπR² = − ^ɛ₀/_ωcos ωt

ω = ɛ₀cos ωt / BπR²

at t = 0,
ω = ɛ₀/ BπR²

Answer: C

Alternative solution:
A = πR²cos ωt
ɛ = − dΦ/dt = − B dA/dt
ɛ = − BπR²d(cos ωt)/dt
ɛ₀ sin ωt = BπR²ωsin ωt
ɛ₀= BπR²ω

ω = ɛ₀/ (BπR²)

Electromagnetism - Faraday’s law

A wire is being wound around a rotating wooden cylinder of radius R. One end of the wire is connected to the axis of the cylinder, as shown in the figure. The cylinder is placed in a uniform magnetic field of magnitude B parallel to its axis and rotates at N revolutions per second. What is the potential difference between the open ends of the wire?

A. 0
B. 2πNBR
C. πNBR²
D. BR²/N
E. πNBR³

(GR9677 #47)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with Φ = NBA

Given:
N revolution per second
B uniform (constant)
A constant

ɛ = − BA dN/dt
= − BπR²N

|ɛ| = πNBR²

Answer: C

Electromagnetism - Conductor

Two long conductors are arranged as shown above to form overlapping cylinders, each of radius r, whose centers are separated by a distance d. Current of density J flows into the plane of the page along the shaded part of one conductor and equal current flows out of the plane of the page along the shaded portion of the other, as shown. What are the magnitude and direction of the magnetic field at point A?

A. (µ₀ ⁄ 2π)πdJ, in the +y-direction
B. (µ₀ ⁄ 2π)d²J ⁄ r, in the +y-direction
C. (µ₀ ⁄ 2π)4d²J ⁄ r, in the −y-direction
D. (µ₀ ⁄ 2π)J²r ⁄ d, in the −y-direction
E. There is no magnetic field at A

(GR9677 #69)

Solution:

Using right-hand rule, the B direction is in the +y direction
→ C, D, E are FALSE

For option (B), if r → 0, B → ∞, B cannot be infinite.

Answer: A

Electromagnetism - Electric and Magnetic Force

A positively charged particle is moving in the xy-plane in a region where there is a non-zero uniform electric magnetic field B in the +z –direction and a non-zero uniform electric field in the +y-direction. Which of the following is a posible trajectory for the particle?

(GR9677 #86)

Solution:

v is in the xy-plane
E is in +y-direction

$\vec{F} = q\vec{E}\rightarrow \vec{F}// \vec{E}$
→ F is in +y-direction
→ particle will be deflected by E in +y-direction

B is in +z-direction

$\vec{F} = q\left (\vec{v}\times \vec{B} \right )$
→ F,v, B orthogonal to each other
→ E and B orthogonal to each other

Particle moving in an orthogonal direction with B will exhibit cyclotron (helix shaped motion).
Answer: B

Electromagnetism - Angular Momentum

Two small pith balls, each carrying a charge q, are attached to the ends of a light rod of length d, which is suspended from the ceiling by a thin torsion-free fiber, as shown in the figure. There is a uniform magnetic field B, pointing straight down, in the cylindrical region of radius R around the fiber. The system is initially at rest. If the magnetic field is turned off, which of the following describes what happens to the system?

It rotates with angular momentum qBR².
It rotates with angular momentum ¼ qBd².
It rotates with angular momentum ½ qBRd.
It does not rotate because to do so would violate conservation of angular momentum.
It does not move because magnetic forces do no work.

(GR9677 #87)

Solution:

B and C are FALSE.
Angular momentum is finite, it cannot go to infinite.
If d → ∞ , angular momentum → ∞

D. FALSE
Since there is external torque, angular momentum is not conserved.

E. FALSE
The system will rotate due to B.

Answer: A:

Calculation:
Faraday's Law: $\varepsilon=-\frac{d \Phi _B}{dt}=\vec{E}\cdot dl$

with Φ = BπR² and dl = 2πd/2 = πd

$\pi R^2 \frac{dB}{dt}=E\pi d\rightarrow E=\frac{\dot{B}R^2}{d}$
$F=qE=\frac{q\dot{B}R^2}{d}$

Torque: $\tau =\sum\vec{r}\times \vec{F}$

Since there is a force contribution from each charge and by the right-hand-rule their cross products with the moment-arm point in the same direction,

$\tau =2 \frac{d}{2}\frac{q\dot{B}R^2}{d}=q\dot{B}R^2$

Torque and Angular momentum:

$\\\sum \tau = \dot{L} \\\rightarrow \dot{L} = q\dot{B}R^2 \\\rightarrow L= qBR^2$

Electromagnetism - Magnetic Field

A coaxial cable has the cross section shown in the figure above. The shaded region is insulated. The regions in which r

a and b

c are conducting. A uniform dc current density of total current I flows along the inner part of the cable (r

a) and returns along the outer part of the cable (b

c) in the direction shown. The radial dependence of the magnitude of the magnetic field, H, is shown by which of the following?

(GR9677 #88)

Solution:

For r = 0 → H = 0
→ D and E are FALSE

For r = c → H = 0 since I_inand I_out cancel each other
→ A and C are FALSE

Answer: B

Electromagnetism - Conductor

A coaxial cable having radii a, b, and c carries equal and opposite currents of magnitude i on the inner and outer conductors. What is the magnitude of the magnetic induction at point P outside of the cable at a distance r from the axis?

A. Zero

B. $\frac{\mu_0 i r}{2\pi a^2}$

C. $\frac{\mu_0 i}{2\pi r}$

D. $\frac{\mu_0 i}{2\pi r} \frac{c^2-r^2}{c^2-b^2}$

E. $\frac{\mu_0 i}{2\pi r} \frac{r^2-b^2}{c^2-b^2}$

(GR9277 #09)

Solution:

The inner and outer conductors carry equal and opposite currents. Thus, the magnitude of the magnetic induction outside the coaxial cable is zero.

Answer: A

Electromagnetism - Faraday's Law

A uniform and constant magnetic field B is directed perpendicularly into the plane of the page everywhere within a rectangular region as shown above. A wire circuit in the shape of a semicircle is uniformly rotated counterclockwise in the plane of the page about an axis A. The axis A is perpendicular to the page at the edge of the field and directed through the center of the straight-line portion of the circuit. Which of the following graphs best approximates the emf ε induced in the circuit as a function of time t?

(GR9277 #57)

Solution:

Faraday's Law: ɛ = − dΦ/dt
with Φ = NBA

Given:
N = 1,
B constant
A increases and decreases uniformly since the wire circuit rotates uniformly
→ the rate of change of A, dA/dt = constant.

ɛ = − BdA/dt = constant

Only graph (A) shows constant ε.
ε changes periodically from positive to negative since only half of area covered in magnetic field.

Answer: A

Special Relativity - Electromagnetic Field

Which of the following statements most accurately describes how an electromagnetic field behaves under a Lorentz transformation?

The electric field transforms completely into a magnetic field.
If initially there is only an electric field, after the transformation there maybe both an electric and magnetic field.
The electric field is unaltered.
The magnetic field is unaltered.
It cannot be determined unless a gauge transformation is also specified.

(GR8677 #22)

Solution:

see analysis: Electromagnetism and Relativity

Answer: A

Electromagnetism - Electric and Magnetic Forces

A charge particle is released from rest in a region where there is a constant electric field and a constant magnetic field. If the two fields are parallel to each other, the path of the particle is a

A. circle
B. parabola
C. helix
D. cycloid
E. straight line

(GR8677 #25)

Solution:

$\vec{F}= \vec{F}_E+\vec{F}_B = q(\vec{E}+\vec{v}\times \vec{B})$

Electric Force: $\vec{F}_E = q\vec{E}$
→ the path of the particle is parallel to the electric field.
→ the velocity of particle has the same direction with the electric field.

Magnetic Force/Lorentz Force: $\vec{F}_B = q(\vec{v}\times \vec{B})$

If $\vec{E} \parallel \vec{B}\rightarrow \vec{v} \parallel \vec{B}\rightarrow \vec{v} \times \vec{B}= 0$
→ no magnetic field contribution
→ the path of the particle is a straight line, parallel to the electric field.

Answer: E

Electromagnetism - Biot Savart's Law

A current i in a circular loop of radius b produces a magnetic field. At a fixed points far from the loop, the strength of the magnetic field is proportional to which of the following combinations of i and b?

A. ib
B. ib²
C. i²b
D. i/b
E. i/b²

(GR8677 #65)

Solution:

Applying the concept of magnetic dipole moment:
μ = a vector quantity associated with the torque exerted by an external magnetic field on a current carrying coil

μ = IA = Iπr² = iπb²

Answer: B

Note: click HERE for complete calculation to show that magnetic field at a fixed distance x far away from a circular loop: $B= \frac{\mu_0 IR^2}{2x^3}$

Electromagnetism - Maxwell's Equation

One of Maxwell’s equation is ∇ ∙ B = 0. Which of the following sketches shows magnetic field lines that clearly violate this equation within the region bounded by the dashed lines?

(GR8677 #79)

Solution:

Gauss' law of magnetism, ∇ ∙ B = 0:
- Total magnetic flux through the gaussian surface is zero
- No magnetic monopole
- B is a constant

(A) and (B) show B is a constant.
Net flux is zero (total lines in = out)
TRUE.

(C) shows magnetic field between two different poles.
Net flux is zero (total lines in = out)
TRUE

(D) shows a magnetic monopole, and
Net flux is not zero (0 lines in, 8 out), and
B is not a constant (spread out).
FALSE

(E) shows magnetic field around a straight wire.
Net flux is zero (no line in and out)
TRUE.

Answer: D

Electromagnetism - Particle's Trajectory

A particle with charge q and momentum p is moving in the horizontal plane under the action of a uniform vertical magnetic field of magnitude B. Measurements are made of the particle's trajectory to determine the “sagitta” s and half-chord length l, as shown in the figure above. Which of the following expressions gives the particle's momentum in terms of q, B, s, and l? (Assume s ≪ l).

A. qBs²/2l
B. qBs²/l
C. qBl/s
D. qBl²/2s
E. qBl²/8s

(GR9677 #89)

Solution:

$\\F=qvB=\frac{mv^2}{R} \\\\qB=\frac{mv}{R}=\frac{p}{R} \rightarrow p=qBR \\\\R^2=l^2+(R-s)^2=l^2+R^2-2Rs+s^2 \\s\ll l \\R^2=l^2+R^2-2Rs \rightarrow l^2=2Rs \rightarrow R=l^2/2s \\\\\rightarrow p=qBR=qBl^2/2s$

Answer: D

Electromagnetism - Magnetic Field

Two long identical bar magnets are placed under a horizontal piece of paper, as shown in the figure above. The paper is covered with iron fillings. When the two north poles are a small distance apart and touching the paper, the iron filings move into pattern that shows the magnetic field lines. Which is the following best illustrates the pattern that results?

(GR0177 #07)

Solution:

(A) and (C) FALSE

(D) FALSE
It could be right if there is a current through the magnets.

(E) FALSE
No magnetic monopoles

Answer: B

Electromagnetism - Lorentz Force

A proton moves in the +z direction after being accelerated from rest through a potential difference V. The proton then passes through a region with a uniform electric field E in the +x direction and a uniform magnetic field B in the +y direction, but the proton’s trajectory is not affected. If the experiment were repeated using a potential difference of 2V, the proton would then be

A. Deflected in the +x-direction
B. Deflected in the -x-direction
C. Deflected in the +y-direction
D. Deflected in the -y-direction
E. Undeflected

(GR0177 #58)

Solution:

Lorentz Force:

Since it is not deflected,

Potential energy transforms into kinetic energy, qV = ½ mv²

Thus, velocity is $v=\sqrt{2qV/m}$

For 2V, velocity is

Therefore,

Particle is deflected in −x direction

Answer: B

Electromagnetism - Conductor

An electromagnetic plane wave, propagating in vacuum, has an electric field given by E = E₀ cos (kx−ωt) and is normally incident on a perfect conductor at x = 0, as shown in figure. Immediately to the left of the conductor, the total electric field E and the total magnetic field B are given by which of the following?

	E	B
A.	0	0
B.	2E₀ cos ωt	0
C.	0	2(E₀/c) cos ωt
D.	2E₀ cos ωt	2(E₀/c) cos ωt
E.	2E₀ cos ωt	2(E₀/c) sin ωt

(GR0177 #61)

Solution:

Perfect conductor:

Its charge and current are distributed on the surface so that external fields can not penetrate it.

Boundary Condition (BC) on the surface of perfect conductor:

E_tangential = 0
E_normal ≠ 0
B_tangential ≠ 0
B_normal = 0

EM wave: E ⊥ B (perpendicular to each other).

If E normal to the surface of conductor, B is tangential.

BC: E_normal ≠ 0

E_initial is reflected by the conductor with the same magnitude but opposite direction.

Total E = E_initial− E_reflected= 0

(B), (D), (E) FALSE

BC: B_tangential ≠ 0

B is not reflected, it's parallel to the surface.

Total B ≠ 0

(A) FALSE

Answer: C

Electromagnetism - Biot Savart's Law

A segment of wire is bent into an arc of radius R and subtended angle θ, as shown in the figure. Point P is at the center of the circular segment. The wire carries current I. What is the magnitude of the magnetic field at P?

A. 0
B. ^μ₀I^θ/₍₂_π₎_²_R

C. ^μ₀I^θ/₄_πR
D. ^μ₀I^θ/₄_πR_²
E. ^μ₀I/₂_θ_R_²

(GR0177 #88)

Solution:

Biot Savart’s Law:

$\begin{aligned} d \vec{B}&=\frac{\mu_0}{4\pi} \frac{Id \vec{l}\times \hat{r}}{r^2} \\\\\hat{r}&=\vec{r}/r \\d \vec{B}&=\frac{\mu_0}{4\pi} \frac{Id \vec{l}\times \vec{r}}{r^3} \end{aligned}$

$\begin{aligned} &Id \vec{l}\perp \vec{R} \rightarrow Id \vec{l}\times\vec{R}=IR dl \\ &d \vec{B}=\frac{\mu_0}{4\pi} \frac{IR dl}{R^3} \\&dl=Rd\theta\rightarrow d \vec{B}=\frac{\mu_0}{4\pi} \frac{IR^2d\theta}{R^3} \\&B=\frac{\mu_0 I\theta}{4\pi R} \end{aligned}$

Answer: C

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